What would the initial rate of formation of HI be if the initial concentrations of the hydrogen gas and iodine gas were each 0.50 M
H2 + I2 --> 2HI
#1) H2 1.0M I2 1.0M 0.20 mol/Ls
#2) H2 1.0M I2 2.0M 0.40 mol/Ls
#3) H2 2.0M I2 2.0M 0.80 mol/Ls
please show steps + any formulas used
The rate law for this reaction is:
Rate = k[H2][I2]
where k is the rate constant.
To find the initial rate of formation of HI, we need to substitute the initial concentrations into the rate law and solve for the rate. In this case, both [H2] and [I2] are 0.50 M.
Rate = k[0.50][0.50]
Rate = 0.25k
We don't know the value of k, so we can't calculate the actual rate. However, we can compare the rates in the different scenarios given:
#1) Rate = 0.25k
#2) Rate = k[1.0][2.0] = 2k
#3) Rate = k[2.0][2.0] = 4k
Comparing #1 and #2, we see that doubling the concentration of I2 doubles the rate of reaction, indicating that the reaction is first order with respect to I2. Comparing #1 and #3, we see that doubling the concentrations of both reactants quadruples the rate of reaction, indicating that the reaction is second order overall.